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a(n) = floor(x^n) where x = 1 + sqrt(5)/2 = A176055.
2

%I #37 Dec 18 2021 23:33:27

%S 1,2,4,9,20,42,90,191,405,857,1816,3848,8150,17263,36564,77445,164031,

%T 347423,735855,1558567,3301098,6991839,14808952,31365865,66433969,

%U 140709405,298027302,631231956,1336970739,2831749467,5997741619,12703420605,26906276616

%N a(n) = floor(x^n) where x = 1 + sqrt(5)/2 = A176055.

%C a(n+1)/a(n) tends to A176055 when n tends towards infinity.

%p seq(floor((1+sqrt(5)/2)^n), n=0..32);

%t a[n_] := Floor[(GoldenRatio + 1/2)^n]; Array[a, 33, 0] (* _Amiram Eldar_, Dec 14 2021 *)

%o (Python)

%o from sympy import floor, sqrt

%o def A350092(n): return floor((1+sqrt(5)/2)**n) # _Chai Wah Wu_, Dec 17 2021

%Y Cf. A176055, A058066 (x*n), A014217 (phi^n).

%K nonn,easy

%O 0,2

%A _Michel Lagneau_, Dec 14 2021